Electronic Journal of Differential Equations, Vol. 2015 (2015), No. 226, pp. 1–11. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu ENERGY DECAY OF A VARIABLE-COEFFICIENT WAVE EQUATION WITH NONLINEAR TIME-DEPENDENT LOCALIZED DAMPING JIEQIONG WU, FEI FENG, SHUGEN CHAI Abstract. We study the energy decay for the Cauchy problem of the wave equation with nonlinear time-dependent and space-dependent damping. The damping is localized in a bounded domain and near infinity, and the principal part of the wave equation has a variable-coefficient. We apply the multiplier method for variable-coefficient equations, and obtain an energy decay that depends on the property of the coefficient of the damping term. 1. Introduction Let n ≥ 2 be an integer. We consider the energy decay for the solution to the Cauchy problem of the wave equation with time-dependent and space-dependent damping, utt − div(A(x)∇u) + u + σ(t)ρ(x, ut ) = 0, u(0, x) = u0 (x), ut (0, x) = u1 (x), x ∈ Rn , t > 0, n x∈R , (1.1) (1.2) n where A(x) = (aij (x)) is a symmetric, positive matrix for each x ∈ R ; σ : R+ → R+ is a non-increasing function of class C 1 . Let Ω and Ω1 be two bounded domains such that Ω1 ⊂ Ω. We assume that ( a(x)ut , x ∈ R n \ Ω1 ρ(x, ut ) = (1.3) a(x)h(ut ), x ∈ Ω1 where h : R → R is a nonlinear continuous nondecreasing function satisfying h(s)s > 0 for all s 6= 0 and a(·) ∈ L∞ (Rn ) is a nonnegative function satisfying x ∈ Ω1 ∪ Ωc . a(x) ≥ ε0 > 0, (1.4) When x ∈ Ω1 , c1 |ut |m ≤ |h(ut )| ≤ c2 |ut |1/m , c3 |ut | ≤ |h(ut )| ≤ c4 |ut |, |ut | ≤ 1, |ut | ≥ 1 where m ≥ 1 and ci > 0 (i = 1, 2, 3, 4) are given numbers. 2010 Mathematics Subject Classification. 35L05, 35L70, 93B27. Key words and phrases. Energy decay; time-dependent and space-dependent damping; localized damping; Riemannian geometry method; variable coefficient. c 2015 Texas State University - San Marcos. Submitted May 28, 2015. Published September 2, 2015. 1 (1.5) (1.6) 2 J. WU, F. FENG, S. CHAI EJDE-2015/226 Since (1.4) implies that the dissipation may vanish in Ω\Ω1 , we call the damping σ(t)ρ(x, ut ) a time-dependent localized damping. Many studies concerning energy decay of wave equations in the bounded domain are available in the literature. We refer the readers to [3, 8, 9, 16]. The author of [8] derived precise energy decay estimates for the initial-boundary value problem to the wave equation with a localized nonlinear dissipation which depended on the time as well as the space variable. The result of [8] is a generalization of the work [9], where the dissipation was independent of the time. A wave equation with timedependent but space-independent damping was investigated in [3] and the decay rate was obtained by solving a nonlinear ODE. For the energy decay of wave equation in the exterior domain or on the whole space, many results have been obtained for the case of constant coefficients, that is A(x) ≡ I (see [1, 2, 4, 10, 11, 12, 17]). The damping in [4, 11, 17] were timeindependent and localized in a sub-domain of Rn . Energy decay of the Cauchy problem for the wave equation with time-dependent damping was studied in [2] where the coefficient of the damping did not depend on the space variable. It is interesting to study the case where the damping is time-dependent and exists on a local domain of Rn . [1] addressed the decay for a wave equation on an exterior domain where the damping is time-dependent and effective only in a ball of Rn . The dimension of the space variable was restricted to be odd since the Lax-Phillips’ scattering theory was used in [1]. Nakao [10] considered the Cauchy problem of wave equation with a dissipation effective outside of a given ball of Rn . However, in the case of variable coefficient, few results about the energy decay of wave equation in the exterior domain or on the whole space can be seen (see[5, 13]). Yao [13] studied the energy decay for the Cauchy problem of the variable-coefficient wave equation with a linear damping. The authors of [5] considered the energy decay of variable-coefficient wave equation with nonlinear damping in an exterior domain. The purpose of this paper is to derive the energy decay for (1.1)-(1.2) with the assumptions (1.3)-(1.6). Our first main goal is to dispense with the restriction A(x) ≡ I. We will use the Riemannian geometry method which was introduced by Yao [14] (see also [15]) to deal with controllability for the wave equation with variable-coefficient principal part and has been viewed as an important method for variable-coefficient models. Our second goal is to generalize the work [2], where the damping exists on the whole space. In this paper, we assume that the damping is localized near the origin point and near infinity and may disappear in a large area. We try to explain how the time-dependence of the damping affect the decay when the dissipation is effective in a localized domain. Energy decay is obtained under some conditions on A(x) and the coefficient of the damping. We introduce a Riemanian metric on Rn by g(x) = A−1 (x) for x ∈ Rn . (1.7) We denote by ∇f and ∇g f the gradients of f in the standard metric of the Euclidean space Rn and in the metric g, respectively. The energy of the model (1.1)-(1.2) is defined by Z 1 (u2 + |∇g u|2g + u2 )dx. (1.8) E(t) = 2 Rn t EJDE-2015/226 ENERGY DECAY 3 where ∇g u = A(x)∇u and |∇g u|2g = h∇g u, ∇g uig = hA(x)∇u, ∇ui = n X aij (x) ij=1 ∂u ∂u . ∂xj ∂xi We refer the readers to [14] and [15] for more information about the metric g(x). We use the following assumption in this article (H1) There is a vector field H on (Rn , g) such that Dg H(X, X) ≥ σ̄|X|2g , X ∈ Rnx , x ∈ Ω̄, (1.9) where σ̄ > 0 is a constant and Dg is the Levi-Civita connection of the metric g. Our main result read as follows. Theorem 1.1. Let (u0 , u1 ) ∈ H 1 (Rn ) × L2 (Rn ). In addition to (H1) assume that σ : R+ → R+ is a non-increasing C 1 function and σ(t) ≥ σ0 > 0 for all t ≥ 0. Then (1.1)-(1.2) admits a unique solution u satisfying 2 m−1 1 E(t) ≤ CE(0) R t , ∀t > 0 if m > 1, σ(s)ds 0 and Z E(t) ≤ CE(0) exp(−ω t σ(s)ds), ∀t > 0 if m = 1, 0 for some ω > 0 and C > 0. 2. Proof of main result To prove our main result we use the following lemma. Lemma 2.1 ([6]). Let E : R+ → R+ be a non-increasing function and φ : R+ → R+ a strictly increasing C 1 function such that φ(0) = 0, φ(t) → +∞ as t → +∞. Assume that there exist r ≥ 0 and ω > 0 such that Z +∞ 1 E 1+r (t)φ0 (t)dt ≤ E r (0)E(S), ω S 0 ≤ S < +∞. Then E(t) ≤ E(0) 1 + r 1/r 1 + ωrφ(t) E(t) ≤ CE(0)e1−ωφ(t) for all t ≥ 0, if r > 0, for all t ≥ 0, if r = 0. To apply Lemma 2.1, we need some estimates on the energy terms, which are based on the identities below. Multiply (1.1) by ut and integrate by parts over Rn with respect to the variable x to obtain Z d E(t) = − σ(t)ρ(x, ut )ut dx. (2.1) dt Rn 4 J. WU, F. FENG, S. CHAI EJDE-2015/226 Let H be a vector field on Rn . Multiply (1.1) by H(u) and integrate by parts over Rn with respect to the variable x to obtain Z Z d ut H(u)dx + Dg H(∇g u, ∇g u)dx dt Rn Rn Z Z 1 + H(u)u dx + (u2 − |∇g u|2g )divHdx (2.2) 2 Rn t n R Z =− σ(t)ρ(x, ut )H(u)dx. Rn Let q ∈ C 1 (Rn ) be a function. Multiply (1.1) by qu and integrate by parts over R with respect to the variable x to obtain Z Z Z Z d 2 2 2 u div A∇q dx + quut dx + q |∇g u|g − ut dx − qu2 dx dt Rn Rn Rn Rn Z (2.3) =− quσ(t)ρ(x, ut )dx. n Rn Proof of Theorem 1.1. Let the vector field H be such that the assumption (H1) ˆ ¯ ˆ holds. Let Ω̂ ⊂ Rn and Ω̂ ⊂ Rn be two bounded domains such that Ω̄ ⊂ Ω̂ ⊂ Ω̂ ⊂ Ω̂. Let ϕ and ψ be two C0∞ functions and 0 ≤ ϕ ≤ 1, 0 ≤ ψ ≤ 1, and ( ( 1, x ∈ Ω̂, 1, x ∈ Ω, ϕ= ψ= (2.4) ˆ c n 0, x ∈ Ω̂ = R \ Ω̂, 0, x ∈ Rn \ Ω̂. Let q0 = div(ϕH) − σ̄ϕ. Replacing H by ϕH in (2.2) and replacing q by q0 in (2.3), 2 respectively, we can obtain two identities. Then we add them up and obtain Z Z d ut ϕH(u) + q0 uut dx + Dg (ϕH)(∇g u, ∇g u)dx dt Rn Rn Z Z + ϕH(u)u dx + σ̄ϕ(u2t − |∇g u|2g )dx n n R ZR Z (2.5) 2 − |u| div(∇g q0 )dx + q0 u2 dx Rn Rn Z Z =− q0 uσ(t)ρ(x, ut )dx − σ(t)ρ(x, ut )ϕH(u)dx. Rn Rn Let k be a large constant determined later. Combining (2.1) with (2.5), we have Z Z d ut ϕH(u) + q0 uut dx + kE 0 (t) + Dg (ϕH)(∇g u, ∇g u)dx dt Rn Rn Z Z Z 2 2 + ϕH(u)u dx + σ̄ϕ(ut − |∇g u|g )dx − |u|2 div(∇g q0 )dx Rn Rn Rn Z + q0 u2 dx (2.6) n R Z Z =− q0 uσ(t)ρ(x, ut )dx − σ(t)ρ(x, ut )ϕH(u)dx Rn Rn Z −k σ(t)ρ(x, ut )ut dx. Rn EJDE-2015/226 ENERGY DECAY 5 Set Z Z Dg (ϕH)(∇g u, ∇g u)dx+ Y (t) = Rn Rn σ̄ϕ(u2t −|∇g u|2g )dx+k Z σ(t)ρ(x, ut )ut dx. Rn Set σ1 = supΩ̂¯ |Dg (ϕH)(X, X)|. Noticing (1.3),(1.4),(1.9) and (2.4), we have the following estimates on Y (t): Z Z Z Dg (ϕ)H(∇g u, ∇g u)dx + σ̄ϕu2t dx − σ̄ϕ|∇g u|2g dx n Ω̂\Ω R \Ω Ω̂\Ω Z Z Z 3k k 2 2 0 σ(t)ut dx + σ̄ut dx + σ(t)ρ(x, ut )ut dx + 4 Rn \Ω 4 Rn Ω Z Z k 2 ε0 σ(t) + σ̄ϕ u2t dx ≥ (−σ1 − σ̄) |∇g u|g dx + 4 n Ω̂\Ω R \Ω Z Z 3k + σ̄u2t dx + σ(t)ρ(x, ut )ut dx . 4 Rn Ω Y (t) ≥ Noticing σ(t) ≥ σ0 > 0, we choose k large enough so that Z Y (t) ≥ Z 2 σ̄|ut | dx − Rn (σ1 + σ̄)|∇g u|2g Ω̂\Ω 3k + 4 kσ0 ε0 4 (2.7) > σ̄. Then we have Z σ(t)ρ(x, ut )ut dx . (2.8) Rn Now we estimate the last term of the right side of (2.8). Z σ(t)ρ(x, ut )ut dx Rn Z Z ≥ ψσ(t)ρ(x, ut )ut dx ψσ(t)ρ(x, ut )ut dx = ˆ Z Ω̂ = ψσ(t)a(x)u2t dx + ψσ(t)ρ(x, ut )ut dx ˆ Ω̂\Ω1 Ω1 Z Z Z = ψσ(t)a(x)u2t dx − σ(t)a(x)u2t dx + ψσ(t)ρ(x, ut )ut dx . n ZR ˆ Ω̂ Ω1 (2.9) Ω1 For the first term on the right of (2.9), noticing σ(t) ≥ σ0 and using (1.4), (2.4) and (2.3) for q = σ0 ψa(x), we have Z ˆ Ω̂ ψσ(t)a(x)u2t dx ≥ σ0 Z Z σ0 ψa(x)u2t dx Z Z d ≥ σ0 a(x)ψuut dx + σ0 ε0 |∇g u|2g dx dt Rn Ω̂\Ω Z Z 2 − σ0 u div A∇ a(x)ψ dx + σ0 a(x)ψu2 dx Rn Rn Z + σ0 a(x)ψσ(t)ρ(x, ut )u dx. ˆ Ω̂ ψa(x)u2t dx = Rn Rn ˆ since supp ψ ⊂ Ω̂ and a(x) ≥ ε0 > 0 for x ∈ Ωc . (2.10) 6 J. WU, F. FENG, S. CHAI Combining (2.8) with (2.9) and (2.10), we have Z kσ ε Z 0 0 2 Y (t) ≥ − (σ1 + σ̄) |∇g u|g dx + σ̄u2t dx 4 Ω̂\Ω Rn Z Z σ0 k h d k σ(t)ρ(x, ut )ut dx + a(x)ψuut dx + 2 n 4 dt Rn Z R Z − u2 div A∇ a(x)ψ dx + a(x)ψu2 dx n n R ZR i kZ σ(t)a(x)u2t dx + a(x)ψσ(t)ρ(x, ut )udx − 4 Ω1 Rn Z k + σ(t)ρ(x, ut )ut dx 4 Ω1 EJDE-2015/226 (2.11) Choose k large enough so that kσ40 ε0 > σ̄ + σ1 . Then it follows from (2.11) and (2.6) that Z Z σ0 k d ut ϕH(u) + q0 uut dx + kE(t) + a(x)ψuut dx dt Rn 4 Rn Z Z Z k k 2 + σ̄ut dx + σ(t)ρ(x, ut )ut dx − σ(t)a(x)ψu2t dx 2 Ω1 4 Ω1 Rn Z Z k σ0 k h + σ(t)ρ(x, ut )ut dx + aψ − div A∇ a(x)ψ) u2 dx 4 Ω1 4 n R (2.12) Z i Z + a(x)ψσ(t)ρ(x, ut )udx + ϕH(u)u dx n n ZR Z R + q0 − div ∇g q0 u2 dx + q0 uσ(t)ρ(x, ut )dx n Rn ZR + σ(t)ρ(x, ut )ϕH(u)dx ≤ 0. Rn We may have equality by setting q = σ̄2 in (2.3). Then add that equality to (2.12) to yield Z Z σ0 k d ut ϕH(u) + q0 uut dx + kE(t) + a(x)ψuut dx dt Rn 4 Rn Z Z σ̄ σ̄ uut dx + |∇g u|2g + u2t + u2 + σ(t)ρ(x, ut )u dx + Rn 2 Rn 2 Z Z k k + σ(t)ρ(x, ut )ut dx − σ(t)a(x)ψu2t dx 2 Ω1 4 Ω1 Z Z k σ0 k h + σ(t)ρ(x, ut )ut dx + aψ − div A∇ a(x)ψ u2 dx (2.13) 4 4 Rn Z Ω1 Z i + a(x)ψσ(t)ρ(x, ut )udx + ϕH(u)u dx Rn Rn Z Z + q0 − div ∇g q0 u2 dx + q0 σ(t)ρ(x, ut )u dx n Rn ZR + σ(t)ρ(x, ut )ϕH(u)dx ≤ 0. Rn EJDE-2015/226 ENERGY DECAY 7 Set Z X(t) = Rn σ0 k ut ϕH(u) + q0 uut dx + kE(t) + 4 Z Z a(x)ψuut dx + Rn Rn σ̄ uut dx. 2 From (2.13) and the definition of E(t), we have σ̄E(t) Z Z σ̄ d σ(t)ρ(x, ut )u dx − σ(t)ρ(x, ut )ϕH(u) dx ≤ − X(t) − dt Rn 2 Rn Z Z k − σ(t)ρ(x, ut )ut dx q0 σ(t)ρ(x, ut )udx − 2 Rn n R Z Z σ0 k h k aψ − div A∇ a(x)ψ u2 dx σ(t)ρ(x, ut )ut dx − − 4 Ω1 4 Rn Z i kZ + a(x)ψσ(t)ρ(x, ut )u dx + σ(t)a(x)ψu2t dx 4 Ω1 Rn Z Z − ϕH(u)u dx − q0 − div ∇g q0 u2 dx Rn Rn k Z d σ(t)ρ(x, ut )u dx − E 0 (t) ≤ − X(t) + σ4 dt 2 n Z ZR Z σ0 kσ2 k +( σ(t)a(x)u2t dx + σ5 |∇g u|g u dx + σ3 ) |u|2 dx + ˆ 4 4 Ω1 Ω̂ Ω̂ Z + σ5 σ(t)|ρ(x, ut )||∇g u|g dx (2.14) Rn where σ2 = sup |aψ − div ∇g (aψ)|, ¯ ˆ x∈Ω̂ σ4 = max σ3 = sup |q0 − div ∇g q0 |, σ0 k|a|∞ σ̄ , , sup |q0 | , 4 2 Ω̂¯ ¯ x∈Ω̂ σ5 = sup |ϕH|. ¯ Ω̂ For the second term, the sixth term and the last term on the right of (2.14), we use Young’s inequality to obtain Z Z σ4 σ(t)ρ(x, ut )udx ≤ σ4 Cε |σ(t)ρ(x, ut )|2 + εu2 dx n Rn ZR (2.15) ≤ σ4 Cε |σ(t)ρ(x, ut )|2 + 2εC4 E(t), Rn Z Z Z ε|∇g u|2g dx + Cε σ5 u2 dx, Rn Ω̂ Z ≤ 2εC5 E(t) + Cε σ5 u2 dx, |∇g u|g udx ≤ σ5 σ5 Ω̂ (2.16) Ω̂ and Z σ5 σ(t)|ρ(x, ut )||∇g u|g dx Z Z ≤ εσ5 |∇g u|2g dx + Cε σ5 |σ(t)ρ(x, ut )|2 dx Rn Rn Z ≤ 2εC5 E(t) + Cε σ5 |σ(t)ρ(x, ut )|2 dx Rn Rn (2.17) 8 J. WU, F. FENG, S. CHAI EJDE-2015/226 where ε > 0 is small and Cε is a constant dependent on ε. Let φ be a function satisfying the conditions in Lemma 2.1. Let r ≥ 0 be a constant determined later. Multiplying (2.14) by E r φ0 and integrating on [S, T ] with respect to t, from (2.15) -(2.17) we have Z T E r+1 (t)φ0 σ̄dt S T Z r 0 ≤− Z 0 T r 0 E φ X (t)dt + (σ4 Cε + σ5 Cε ) S Z T E r+1 φ0 dt − + (2εσ4 + 4εσ5 ) S k 2 Z Z Z |σ(t)ρ(x, ut )|2 dx dt E φ Rn S T E r φ0 E 0 dt (2.18) S Z T kσ0 σ2 u2 dx dt + σ3 + C ε σ5 ) E r φ0 ˆ 4 Ω̂ S Z Z k T r 0 E φ σ(t)a(x)u2t dx dt. + 4 S Ω1 +( Using a similar argument as in [13] and [7], we can prove that if T − S is large enough, then Z T E r φ0 S Z ˆ Ω̂ u2 dx dt ≤ Cη T Z E r φ0 Z |σ(t)ρ(x, ut )|2 dx Rn S Z Z σ(t)u2t dx dt + η + Ω1 (2.19) T E r+1 (t)φ0 dt S holds for any η > 0. Rt For the rest of this article, let φ(t) = 0 σ(s)ds. It is clear that φ(t) satisfies the conditions of Lemma 2.1. Noticing that |X(t)| ≤ CE(t) and E 0 (t) ≤ 0, we have by integrating by parts with respect to t, Z T − E r φ0 X 0 (t)dt S T Z T = −E (t)φ (t)X(t) + (E r φ0 )0 X(t)dt S S Z T Z r+1 r−1 0 0 ≤ Cσ(0)E (S) + |rE E φ X(t)|dt + r 0 S ≤ Cσ(0)E r+1 (S) + Cσ(0) Z T S rE r (−E 0 )dt + Z T |E r φ00 X(t)|dt (2.20) S T E r+1 (−φ00 )dt S ≤ CE r+1 (S). We also have − k 2 Z T E r φ0 E 0 dt ≤ CE r+1 (S). (2.21) S Now we estimate the second term on the right of (2.18). Set Ω+ 1 = {x ∈ Ω1 ; |ut | ≥ + − 1} and Ω− = {x ∈ Ω ; |u | < 1}. Then Ω = Ω ∪ Ω . Noticing (1.5), using Hölder 1 t 1 1 1 1 EJDE-2015/226 ENERGY DECAY 9 inequality and Young’s inequality, we have Z T Z E r φ0 |σ(t)ρ(x, ut )|2 dx dt Ω− 1 S T Z r 0 = Z σ 2 (t)|a(x)h(ut )|2 dx dt E φ Ω− 1 S Z T E r φ0 σ 2 (t) ≤ c2 Z Ω− 1 S Z a2 (x) h(ut )ut T Z E r φ0 = c2 Ω− 1 S T Z 2 2− m+1 σ(t)a(x)h(ut )ut 2 m+1 Z 2 E r φ0 σ(t)2− m+1 ≤ C|a|∞ Ω− 1 S 2 2− m+1 ≤ C|a|∞ 2 m+1 h σ 2 (0) ε1 T Z dx dt 2 2− m+1 a(x)σ(t) dx dt (2.22) 2 m+1 dt σ(t)a(x)h(ut )ut dx 2 E r φ0 σ(t)− m+1 m+1 m−1 dt S T Z Z + Cε1 S Ω− 1 Now, we set r = m−1 2 . T Z i σ(t)a(x)h(ut )ut dx dt . Thus, we have from (2.22) and (2.1) Z r 0 |σ(t)ρ(x, ut )|2 dx dt E φ Ω− 1 S 2− 2 C|a|∞ m+1 σ 2 (0) h ≤ Z (2.23) T ε1 E m+1 2 0 φ dt + Cε1 E(S) . S Noticing (1.6) and (2.1), it is easy to obtain Z Z T Z |σ(t)ρ(x, ut )|2 dx dt ≤ c4 σ(0)|a|∞ E r φ0 Ω+ 1 S T E r φ0 (−E 0 (t))dt S ≤ CE r+1 (2.24) (S). On the other hand, it is easy to obtain Z T Z E r φ0 |σ(t)ρ(x, ut )|2 dx dt ≤ CE r+1 (S) (2.25) Rn \Ω1 S since ρ(x, ut ) = a(x)ut when x ∈ Rn \ Ω1 . From (2.23)-(2.25), we have Z T Z E r φ0 |σ(t)ρ(x, ut )|2 dx dt Rn S ≤ 2− 2 Cε1 |a|∞ m+1 σ 2 (0) Z (2.26) T E m+1 2 0 φ dt + Cε1 E(S) + CE r+1 (S). S Similarly, for the last term on the right side of (2.18), we have Z T Z E r φ0 σ(t)a(x)u2t dx dt S Ω1 m−1 m+1 Z (2.27) T ≤ Cε1 |a|∞ σ(0) E S m+1 2 0 φ dt + Cε1 E(S) + CE r+1 (S). 10 J. WU, F. FENG, S. CHAI EJDE-2015/226 At this point, we choose ε and η small enough so that kσ0 σ2 σ̄ 2εσ4 + 4εσ5 + η( + σ3 + Cε σ5 ) < . 4 2 Then using (2.19)-(2.21), (2.18) becomes Z σ̄ T r+1 E (t)φ0 dt 2 S kσ0 σ2 ≤ CE r+1 (S) + σ4 Cε + σ5 Cε + Cη ( + σ3 + Cε σ5 ) 4 Z Z T Z T k r 0 2 E φ |σ(t)ρ(x, ut )| dx dt − × E r φ0 E 0 dt 2 S S Rn Z Z T r 0 k kσ0 σ2 + E φ σ(t)a(x)u2t dx dt. + Cη ( + σ3 + Cε σ5 ) 4 4 S Ω1 (2.28) Once ε and η are fixed, we pick ε1 small enough so that m−1 σ̄ kσ0 σ2 k m+1 + σ3 + C ε σ5 ) ≤ . 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